A150.0-g sample of a metal at 75.08c is added to 150.0 g h2o at 15.08c. the temperature of the water rises to 18.38c. calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

1) Data:

Mass of metal: 150.0 g
Initial temperature of metal: 75.08 °C
Specific heat: Cs

Mass of water: 150.0 g
Initial temperature of water: 15.08°C

Final temperaute: 18.38°C

2) Analysis:

- The final temperature of the system (metal/water), when the equilibrium is reached, is uniform = 18.38°C

- The heat gain by the waer is equal to the heat lost by the metal.

- The specific heat of water is 1.00 cal / g °C

3) Fórmulas

Heat lost or gained = mass * Cs * ΔT

4) Solution

Heat gained by the water = 150.0 g * 1 cal / g °C * [18.38 - 15.08]°C

Heat lost by the metal = 150.0 g * Cs * [75.08 - 18.38]°C

Heat gained by the water = heat gained by the metal

150.0 g * 1 cal / g °C * [18.38 - 15.08]°C = 150.0 g * Cs * [75.08 - 18.38]°C

[18.38 - 15.08]  cal /g  °C = Cs [75.08 - 18.38]

Cs = 3.3 / 56.7 = 0.058 Cal / g °C < answer

You can pass that to joules / g °C in case you need in that SI unit.

Idid the calculation and i get 875,000,000 which can be written as 8.75 x 10^6?